SATHEE: Kinematics Question 412

Kinematics Question 412

Question: A ball is thrown vertically upwards. It was observed, at a height h twice with a time interval $Δ t$ . The initial velocity of the ball is

Options:

A) $\sqrt{8 g h + g^{2} {(Δ t)}^{2}}$

B) $\sqrt{8 g h + {(\frac{g Δ t}{2})}^{2}}$

C) $\frac{1}{2} \sqrt{8 g h + g^{2} {(Δ t)}^{2}}$

D) $\sqrt{8 g h + 4 g^{2} {(Δ t)}^{2}}$

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Answer:

Correct Answer: C

Solution:

$h=ut - \frac{1}{2} a t^{2}$
$\Rightarrow g t^{2} - 2 u t + 2 h = 0$

Solving for t we get $t_{1} + t_{2} = 2 u / g$

$t_{1} \times t_{2} = 2 h / g$

$so, Δ t= | t_{1} - t_{2} | = {(t_{1} + t_{2})}^{2} - 4 t_{1} t_{2}$

Putting value we get $u= \frac{1}{2} \sqrt{8 h g + g^{2} Δ t^{2}}$

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Kinematics Question 412

Question: A ball is thrown vertically upwards. It was observed, at a height h twice with a time interval Δt . The initial velocity of the ball is

Options:

Answer:

Solution:

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Question: A ball is thrown vertically upwards. It was observed, at a height h twice with a time interval $Δ t$ . The initial velocity of the ball is